Yet Another Haskell Tutorial/Language advanced/Solutions

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Yet Another Haskell Tutorial
Getting Started
Language Basics (Solutions)
Type Basics (Solutions)
IO (Solutions)
Modules (Solutions)
Advanced Language (Solutions)
Advanced Types (Solutions)
Monads (Solutions)
Advanced IO

Sections and Infix Operators[edit | edit source]

Local Declarations[edit | edit source]

Partial Application[edit | edit source]

Function func3 cannot be converted into point-free style. The others look something like:

func1 x = map (*x)

func2 f g = filter f . map g

func4 = map (+2) . filter (`elem` [1..10]) . (5:)

func5 = flip foldr 0 . flip . curry 

You might have been tempted to try to write func2 as filter f . map, trying to eta-reduce off the g. In this case, this isn't possible. This is because the function composition operator (.) has type (b -> c) -> (a -> b) -> (a -> c). In this case, we're trying to use map as the second argument. But map takes two arguments, while (.) expects a function which takes only one.

Pattern Matching[edit | edit source]

Guards[edit | edit source]

Instance Declarations[edit | edit source]

The Eq Class[edit | edit source]

The Show Class[edit | edit source]

Other Important Classes[edit | edit source]

The Ord Class[edit | edit source]

The Enum Class[edit | edit source]

The Num Class[edit | edit source]

The Read Class[edit | edit source]

Class Contexts[edit | edit source]

Deriving Classes[edit | edit source]

Datatypes Revisited[edit | edit source]

Named Fields[edit | edit source]

More Lists[edit | edit source]

Standard List Functions[edit | edit source]

List Comprehensions[edit | edit source]

Arrays[edit | edit source]

Finite Maps[edit | edit source]

Layout[edit | edit source]

The Final Word on Lists[edit | edit source]

We can start out with a recursive definition:

and [] = True
and (x:xs) = x && and xs

From here, we can clearly rewrite this as:

and = foldr (&&) True

We can write this recursively as:

concatMap f [] = []
concatMap f (x:xs) = f x ++ concatMap f xs

This hints that we can write this as:

concatMap f = foldr (\a b -> f a ++ b) []

Now, we can do point elimination to get:

     foldr (\a b -> f a ++ b) []
==>  foldr (\a b -> (++) (f a) b) []
==>  foldr (\a -> (++) (f a)) []
==>  foldr (\a -> ((++) . f) a) []
==>  foldr ((++) . f) []